Complete the equation. $\dfrac32+ \dfrac32+ \dfrac32~=~$
Answer: Let's figure out what $\dfrac{3}{2}+ \dfrac{3}{2} + \dfrac{3}{2}$ equals. $\dfrac{0}{2}$ $\dfrac{3}{2}$ $\dfrac{6}{2}$ $\dfrac{9}{2}$ $\llap{{+}}\!\frac{3}{2}$ $\llap{{+}}\!\frac{3}{2}$ $\llap{{+}}\!\frac{3}{2}$ $\dfrac{3}{2}+ \dfrac{3}{2} + \dfrac{3}{2} = \dfrac{9}{2}$ Now, let's figure out how many times we add $\dfrac{1}{2}$ to make $\dfrac{9}{2}$. $\dfrac{0}{2}$ $\dfrac{1}{2}$ $\dfrac{2}{2}$ $\dfrac{3}{2}$ $\dfrac{4}{2}$ $\dfrac{5}{2}$ $\dfrac{6}{2}$ $\dfrac{7}{2}$ $\dfrac{8}{2}$ $\dfrac{9}{2}$ $\llap{{+}}\!\frac{1}{2}$ $\llap{{+}}\!\frac{1}{2}$ $\llap{{+}}\!\frac{1}{2}$ $\llap{{+}}\!\frac{1}{2}$ $\llap{{+}}\!\frac{1}{2}$ $\llap{{+}}\!\frac{1}{2}$ $\llap{{+}}\!\frac{1}{2}$ $\llap{{+}}\!\frac{1}{2}$ $\llap{{+}}\!\frac{1}{2}$ $=\overbrace{{\dfrac1{2}} +{\dfrac1{2}} +{\dfrac1{2}} + {\dfrac1{2}} +{\dfrac1{2}} +{\dfrac1{2}} + {\dfrac1{2}} + {\dfrac1{2}} + {\dfrac1{2}}}^{{9}\text{ halves}} $ $=\dfrac{{9}\times{1}}{{4}}$ $\dfrac{3}{2}+ \dfrac{3}{2}+ \dfrac{3}{2} = 9 \times \dfrac12$